Let \begin{align} \lambda'(p)=-c\frac{1}{(p+2)^\frac{3}{2}}\frac{\Gamma\left(\frac{3p}{4}-\frac{1}{2}\right)}{\Gamma\left(\frac{3p}{4}\right)}\left(\frac{3}{2(p+2)}+\frac{3}{4}\left(\psi(\frac{3p}{4})-\psi(\frac{3p}{4}-\frac{1}{2}) \right) \right) \end{align} where $\psi$ is the digamma function.
By Gautschi's inequality \begin{equation} \frac{\Gamma\left(\frac{3p}{4}-\frac{1}{2}\right)}{\Gamma\left(\frac{3p}{4}\right)}>\left(\frac{3p}{4}\right)^{-\frac{1}{2}} \end{equation} It is stated that the fact that $\psi$ is increasing on $(0,\infty)$, it follows \begin{equation*} |\lambda'(p)|\geq \frac{3c}{2}(p+2)^{-\frac{5}{2}}\left(\frac{3p}{4}\right)^{-\frac{1}{2}}\gtrsim 1, \end{equation*} uniformly for $p\in \left(\frac{4}{3},4\right)$.
I don't understand in the end how I use the fact that $\psi$ is increasing to lower bound $|\lambda'(p)|$
I believe this is a direct consequence of the increasing nature of $\psi(p)$
If $\psi(p)$ is increasing, then by definition, we must have
$$\psi(m)-\psi(n) \ge 0$$
If $m \ge n$
This will imply that
$$\frac{3}{4} \cdot [\psi(\frac{3p}{4})-\psi(\frac{3p}{4}-\frac{1}{2})] \ge 0$$
The rest follows directly as a consequence of this and Gautschi's inequality.
$$\lambda'(p ) \ge -c \cdot \frac{1}{(p+2)^{\frac{3}{2}}} \cdot (\frac{3p}{4})^{-\frac{1}{2}} \cdot \frac{3}{p+2}$$